Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $z = \dfrac{-2p + 14}{p + 3} \div \dfrac{-4p^2 + 28p - 40}{-3p^2 + 6p + 45} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $z = \dfrac{-2p + 14}{p + 3} \times \dfrac{-3p^2 + 6p + 45}{-4p^2 + 28p - 40} $ First factor out any common factors. $z = \dfrac{-2(p - 7)}{p + 3} \times \dfrac{-3(p^2 - 2p - 15)}{-4(p^2 - 7p + 10)} $ Then factor the quadratic expressions. $z = \dfrac {-2(p - 7)} {p + 3} \times \dfrac {-3(p - 5)(p + 3)} {-4(p - 5)(p - 2)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac {-2(p - 7) \times -3(p - 5)(p + 3) } {(p + 3) \times -4(p - 5)(p - 2) } $ $z = \dfrac {6(p - 5)(p + 3)(p - 7)} {-4(p - 5)(p - 2)(p + 3)} $ Notice that $(p - 5)$ and $(p + 3)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac {6\cancel{(p - 5)}(p + 3)(p - 7)} {-4\cancel{(p - 5)}(p - 2)(p + 3)} $ We are dividing by $p - 5$ , so $p - 5 \neq 0$ Therefore, $p \neq 5$ $z = \dfrac {6\cancel{(p - 5)}\cancel{(p + 3)}(p - 7)} {-4\cancel{(p - 5)}(p - 2)\cancel{(p + 3)}} $ We are dividing by $p + 3$ , so $p + 3 \neq 0$ Therefore, $p \neq -3$ $z = \dfrac {6(p - 7)} {-4(p - 2)} $ $ z = \dfrac{-3(p - 7)}{2(p - 2)}; p \neq 5; p \neq -3 $